#-(and) "

P64 (**) Layout a binary tree (1)

    Given a binary tree as the usual Prolog term t(X,L,R) (or nil). As
    a preparation for drawing the tree, a layout algorithm is required
    to determine the position of each node in a rectangular
    grid. Several layout methods are conceivable, one of them is shown
    in the illustration below.

    In this layout strategy, the position of a node v is obtained by the following two rules:

      □ x(v) is equal to the position of the node v in the inorder sequence
      □ y(v) is equal to the depth of the node v in the tree

    In order to store the position of the nodes, we extend the Prolog
    term representing a node (and its successors) as follows:

    % nil represents the empty tree (as usual)
    % t(W,X,Y,L,R) represents a (non-empty) binary tree with root W
    %  \"positioned\" at (X,Y), and subtrees L and R

    Write a predicate layout-binary-tree/2 with the following specification:

    % layout-binary-tree(T,PT) :- PT is the \"positioned\" binary tree
    % obtained from the binary tree T. (+,?)

    Test your predicate in an appropriate way.

(load "p54a")

;; To add the coordinate, we create a new structure, which inherits
;; from the binary-tree structure, so we can reuse that abstraction.
;; However, including structures will make the new fields added at the
;; end of it.  The order of the fields should be immaterial (only that
;; we don't use true structures, but lists, so the new fields are
;; added at the end of the lists, compared to the included list
;; structures).

(defstruct (layout-binary-tree (:include binary-tree)
                               (:type list))
  x y)

(defun binary-tree-to-layout-binary-tree (tree)
Return a layout-binary-tree homologue to node.
  (if (binary-tree-empty-p tree)
       :label (binary-tree-label tree)
       :left  (binary-tree-to-layout-binary-tree (binary-tree-left  tree))
       :right (binary-tree-to-layout-binary-tree (binary-tree-right tree)))))

;; To layout the binary tree, we will do it in two steps.   First we
;; make the layout tree, and setting the y field to the depth of each
;; node.  Then we execute a infix walk of the new tree updating the x
;; field of each node.

(defun layout-node-depth (node depth)
Return a layout-binary-tree homologue to node, with the ordinates of
each node set to their depth.
  (if (binary-tree-empty-p node)
       :label (binary-tree-label node)
       :y depth
       :left  (layout-node-depth (binary-tree-left  node) (1+ depth))
       :right (layout-node-depth (binary-tree-right node) (1+ depth)))))

;; Note, incf is a prefix increment, it returns the new-value.
;; Therefore it is easier to start with the predecessor of the first
;; value, and to finally return the last value used.  One could define
;; a postfix increment operator to easily write the code using the
;; other convention.

(defun layout-node-abscissa/inorder (node abscissa)
Sets the abscissa of each node in the subtree NODE to a sequence of
values starting from (1+ ABSCISSA) for the left-most node.
Returns the last abscissa used.
    ((binary-tree-empty-p (binary-tree-left node))
     (setf (layout-binary-tree-x node) (incf abscissa)))
    ((binary-tree-empty-p (binary-tree-right node))
     (setf (layout-binary-tree-x node)
           (layout-node-abscissa/inorder (binary-tree-left node)
     (layout-node-abscissa/inorder (binary-tree-right node)
                                   (1+ (setf (layout-binary-tree-x node)
                                             (layout-node-abscissa/inorder (binary-tree-left node)

(defun layout-binary-tree/inorder/depth (tree)
  (let ((lobt (layout-node-depth tree 0)))
    (layout-node-abscissa/inorder lobt 0) ; starts from 1; use -1 to start from 0.

;; (layout-binary-tree/inorder/depth  (complete-binary-tree 5))

;;;; THE END ;;;;